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Dear StackExchange community,

I hope this is the right place for my question. This question is motivated by the photographical topic, but is about a purely mathematical issue.

In astrophotography, a common problem is the "trailing" of stars in pictures that have been taken with too long exposure times. Point-shaped stars are pictured as curved lines. This of course originates in the earth's rotation around itself.

To provide a guideline for the longest exposure time possible (in the following: t), there is the "300 rule":

t = 300 / f

with the focal length f.

As far as I can say, this rule is purely empiric. I wonder how a rule like this could be deduced in a mathematically accurate way. My first approach was the following:

  1. The earth's velocity depends on angular velocity ω and radius r, which are

    ω = 2π / (86400 s)

    r = r_earth * cos(latitude)

    with the mean earth radius r_earth.

  2. The critical distance which leads to trailing depends on the camera sensor's dimensions and resolution. If, for example, a maximum trailing of one pixel horizontally would be allowed (i. e. one star affects two neighbouring pixels), this critical distance would be

    s = w / n

    with the sensor's width w and the number of pixels per line n.

  3. All in all, this leads to the equation

    s = ω * r * t ⇔ t = s / (ω * r)

Of course, this is a very bad oversimplification as I totally ignore the artificial manipulation of r via the lens with a certain focal length. However, I thought I would get too large numbers in this approach. Instead, with my sensor's and location's data I get t = 1.7 * 10⁻⁸ s. As you see in the equation above, even if I accept a desastrous amount of 1000 neighbouring pixels that are affected by the same one-pixel star, I do not compute a realistic exposure time.

So, where is the error? And how would I correctly integrate the focal length of the lens in my approach?

Thank you for your suggestions!

  • I’m voting to close this question because it better belongs to Astronomy SE. – ZeroTheHero Apr 20 '20 at 05:48
  • @ZeroTheHero That's not really a reason to close. Being out of this site's scope is a reason to close a question, but being within another site's scope is not. (And in fact, most questions which would be on topic on [astronomy.se] are also on topic here, although astrophotography might be one of the differences.) – David Z Apr 20 '20 at 06:03
  • @ZeroTheHero I'm happy to move this question to Astronomy StackExchange, I just didn't know where to put it. If there is no protest concerning this in a day's time I will delete this question and put it on the site mentioned. – edfrank Apr 20 '20 at 13:20
  • @edfrank obviously my suggestion to close and move has not caught on. On the other hand your post only has 20 views in 19 hrs... I still think the expertise would be with AstroSE but I could be wrong. – ZeroTheHero Apr 20 '20 at 13:21
  • In the calculation, the earth's radius is not relevant. Instead think about how long it takes for the direction you are looking in to move through an angle, which at the focal point of the lens, gives an image displacement that is less than your critical distance (say 1 pixel). The distances you need are thus the focal length of the the lens and the width of a pixel. – Dr Chuck Apr 20 '20 at 17:19
  • @DrChuck Sounds like a major misunderstanding on my side. Could you please elaborate on that? Why doesn't the time for the image displacement depend on the surface velocity of the earth? – edfrank Apr 21 '20 at 13:48
  • @edfrank: for simplicity let's assume we are looking at a star on the celestial equator. Its angular rotation rate (radian/s) is 2pi/86400. Further assume a lens with focal length of 300mm, and a pixel size of 5microns. An isoscalese triangle with base 5 microns and sides 300mm makes an angle of 5/300000. How long does it take the earth to rotate through that angle? (5/300000)/(2pi/86400) =0.23 seconds. So if your criterion is a star trail of 1 pixel, the exposure should be no more than 0.23 seconds. The formula you cite would give 1 second, which corresponds to 3 pixels - still small – Dr Chuck Apr 21 '20 at 17:14
  • And yes I should use Latex but it's after my time. – Dr Chuck Apr 21 '20 at 17:15

0 Answers0